# Solution

## Transcription

Solution
```Math 243
Spring 2015
(Practice) Midterm 2
4/2/2015
Time Limit: 50 minutes
Name:
• No calculators or notes are allowed.
• One side of each sheet is blank and may be used as scratch paper.
Grade Table (for instructor use only)
Question Points Score
1
15
2
20
3
20
4
15
5
30
Total:
100
Math 243
(Practice) Midterm 2 - Page 2 of 8
Common trig function values
x
sin x
cos x
tan x
0
0
1
0
π
6
1
2
π
4
√
3
2
√1
3
√1
2
π
3
√1
2
√
3
2
1
2
1
√
3
π
2
1
0
undefined
Trig identities
• sin2 x + cos2 x = 1
• tan2 x + 1 = sec2 x
• sin2 x =
1−cos 2x
2
• cos2 x =
1+cos 2x
2
• sin 2x = 2 sin x cos x
4/2/2015
The cross product
If ~u = ha, b, ci and ~v = hp, q, ri, then
~u × ~v = hbr − cq, cp − ar, aq − bpi.
Differential geometry
d~r
~v
• Tˆ =
=
ds
|~v |
ˆ
ˆ
ˆ = 1 dT = 1 dT
• N
dTˆ dt
κ ds
dt dTˆ dTˆ 1
• κ= =
ds |~v | dt Z
b
|~v (t)| dt
• arclength =
a
Math 243
(Practice) Midterm 2 - Page 3 of 8
4/2/2015
1. (15 points) Find an equation for the plane that is parallel to the vectors h2, 1, −1i and
h0, −2, 3i and that passes through the point (1, 2, 3).
Solution: First take the cross product of the two vectors to find a vector that is
normal to the plane:
h2, 1, −1i × h0, −2, 3i = h1, −6, −4i.
The equation for the plane is:
x − 6y − 4z = h1, −6, −4i · h1, 2, 3i = 23.
Math 243
(Practice) Midterm 2 - Page 4 of 8
4/2/2015
2. Consider the planes 3x + 6z = 1 and 2x + 2y − z = 3.
(a) (10 points) What is the angle between these planes?
Solution: We have to find the angle between the normal vectors h3, 0, 6i and
h2, 2, −1i. The dot product of these two vectors is 0, so they are orthogonal.
The planes are at a right angle to each other.
(b) (10 points) Find a parametrization for the line of intersection of these planes.
Solution: First, notice that the point (1/3, 7/6, 0) is on the line (set z = 0).
The cross product of the two normal vectors, h−12, 15, 6i is parallel to the line.
So we have
~r(t) = h1/3, 7/6, 0i + th−12, 15, 6i.
Math 243
(Practice) Midterm 2 - Page 5 of 8
4/2/2015
3. A particle moves with constant acceleration ~a = h0, 0, −1i, starting at the point (0, 0, 1)
and with an initial velocity of h1, 1, 0i.
(a) (10 points) When and where does the particle fall below the plane z = 0?
Solution: First,
~v (t) = h1, 1, −ti,
so
t2
.
~r(t) = t, t, 1 −
2
√
2
The z coordinate is 0 when t2 = 1, i.e., when t = 2.
(b) (10 points) Find a vector equation for the tangent line to the particle’s trajectory
at the point where it hits the plane z = 0.
√
√
√
√ √
Solution: ~v ( 2) = h1, 1, −
2i,
and
~
r
(
2)
=
h
√ √
√ 2, 2, 0i, so the equation of
~
the tangent line is l(s) = h 2, 2, 0i + sh1, 1, − 2i.
Math 243
(Practice) Midterm 2 - Page 6 of 8
4/2/2015
4. (15 points) A particle’s position at time t is given by ~r(t) = h1, 3t2 , t3 i. Find the length
of the arc traced out by the particle between time t = 0 and t = 1.
Solution:
Z
1
Z
1
|~v (t)| dt =
0
|~v (t)| dt
0
Z
1
=
√
36t2 + 9t4 dt
0
Z
=
1
√
3t 4 + t2 dt
0
1
= (4 + t2 )3/2 0
= 53/2 − 8
Math 243
(Practice) Midterm 2 - Page 7 of 8
4/2/2015
5. A particle’s position at time t is given by ~r(t) = h1, et cos t − 1, et sin ti.
(a) (10 points) Find the particle’s speed at time t.
Solution:
~v (t) = h0, et cos t − et sin t, et sin t + et cos ti = et h0, cos t − sin t, cos t + sin ti.
So the speed is
|~v (t)| =
√
2et .
(b) (10 points) Find Tˆ(t), the unit tangent vector to the particle’s trajectory at time
t.
Solution:
~v (t)
1
Tˆ(t) =
= √ h0, cos t − sin t, cos t + sin ti.
|~v (t)|
2
Math 243
(Practice) Midterm 2 - Page 8 of 8
4/2/2015
(c) (5 points) Find the curvature κ(t).
Solution:
1 dTˆ 1
κ(t) =
= √ t
|~v (t)| dt 2e
1
√ h0, − sin t − cos t, − sin t + cos ti = √1 .
2
2et
ˆ (t).
(d) (5 points) Find the unit normal vector N
Solution:
ˆ (t) = √1 h0, − sin t − cos t, − sin t + cos ti.
N
2
```